Let CD be the building such that CD = 100 m.
Let AB be the tower of height h metre. It is given that the angles of depression of the top A and the bottom B of the lower AB are 45° and 60° respectively.
i.e., ∠EAC = 45° and ∠DBC = 60°
Let BD = AE = x
In right triangle AEC, we have
In right triangle BDC, we have
Comparing (i) and (ii), we get
Hence,Height of tower (AB) = 42.26 m.
i.e., ∠ACB = β
and ∠ADB = α
Let BC = x metres
In right triangle ABC, we have
In right triangle ABD, we have
Comparing (i) and (ii), we get
Hence, the height of the tower be 180 mts.
AD and BE are two towers. The angle of depression of 1st tower (AD), when seen from the top of 2nd tower (BE) is 30°.
i.e., ∠CDE = 30°.
It is given that
AD = 60 m
and AB = CD = 140 m.
Let height of 2nd tower BE be h metres.
In right triangle CDE, we have
Hence, the height of the second tower is 140.83 m.
Let A be the position of boy D, be the position of girl and F be the position of kite, such that AF - 100 m; CD = 20 m; ∠BAF = 30° and ∠EDF = 45°
In right triangle ABF, we have
Let DF be the length of the second kite.
Now, in right triangle DEF, we have
Hence, the length of te string = 
Hence, the usual speed of the aircraft be x km/hr = 750 km/hr.
Let C be position of the man. AB be the water level, and BH be the hill. The angles of elevation of the top and depression of foot from the deck of the ship be 60° and 30° respectively.
i.e., ∠DCH = 60° and ∠BCD = 30°
Let HD = x m
In right triangle CDH, we have
In right triangle CDB, we have
Hence, distance of the ship from the hill 
Comparing (i) and (ii), we get
Now, total height of the hill
= BD + DH = 10 + x
= 10 + 30 = 40 m
Hence, height of the hill = 40 m.
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